# integral of Lebesgue Measurable function defined on finite measure.

Let $f(x)$ be a nonnegative Lebesgue measurable function on $[a,b]$ and let $E_n=\{x : f(x) \ge n \}$. How to prove that $f$ is integrable if and only if $$\sum_{n=1}^{\infty} \mu(E_n)<\infty.$$

I tried to use chebyshev inequality but It seems its not work ( Let $f$ be a nonnegative measurable functions on $E$. Then for any $\lambda >0 \, \mbox{we have } \mu(\{x\in E : f(x) \ge \lambda \}) \le \frac{1}{\lambda} \int_{E} f$ ). Since the Chyebyshev inequality only give me a lower bound to the integral , which in this case not useful.

• Already asked several times. – Did Dec 26 '14 at 20:22

Note that $$E_k-E_{k+1}=\{x:k\le f(x)\lt k+1\}$$ and since $E_{k+1}\subset E_k$, we have $\mu(E_k-E_{k+1})=\mu(E_k)-\mu(E_{k+1})$. Therefore, \begin{align} \int_{f(x)\ge n}f(x)\,\mathrm{d}x &\ge\sum_{k=n}^\infty k(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{k=n}^\infty\sum_{j=1}^k(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{j=1}^\infty\sum_{k=\max(j,n)}^\infty(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{j=1}^{n-1}\sum_{k=n}^\infty(\mu(E_k)-\mu(E_{k+1})) +\sum_{j=n}^\infty\sum_{k=j}^\infty(\mu(E_k)-\mu(E_{k+1}))\\ &=(n-1)\mu(E_n)+\sum_{j=n}^\infty\mu(E_j) \end{align} Thus, if $f$ is integrable, $\sum\limits_{j=n}^\infty\mu(E_j)\lt\infty$.
Furthermore, \begin{align} \int_{f(x)\ge n}f(x)\,\mathrm{d}x &\le\sum_{k=n}^\infty(k+1)(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{k=n}^\infty\sum_{j=0}^k(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{j=0}^\infty\sum_{k=\max(j,n)}^\infty(\mu(E_k)-\mu(E_{k+1}))\\ &=\sum_{j=0}^{n-1}\sum_{k=n}^\infty(\mu(E_k)-\mu(E_{k+1})) +\sum_{j=n}^\infty\sum_{k=j}^\infty(\mu(E_k)-\mu(E_{k+1}))\\ &=nE_n+\sum_{j=n}^\infty\mu(E_j) \end{align} Thus, if $\sum\limits_{j=n}^\infty\mu(E_j)\lt\infty$, then $f$ is integrable.
Hint: Let $I_n$ be the indicator function of $E_n$. What inequalities relate $\sum_{n=1}^\infty I_n$ and $f$?
Let $E_\alpha := \{x : f(x) \ge \alpha\}$, for all $\alpha \ge 0$. Let $n \in \Bbb N$. For each $\alpha \in [n-1,n]$, $E_n \subset E_\alpha \subset E_{n-1}$ and thus $\mu(E_n) \le \mu(E_\alpha) \le \mu(E_{n-1})$. Hence $$\mu(E_n) \le \int_{n-1}^n \mu(E_\alpha) \, d\alpha \le \mu(E_{n-1}).$$ It follows that $$\sum_{n = 1}^N \mu(E_n) \le \int_0^N \mu(E_\alpha)\, d\alpha \le \mu(E_0) +\sum_{n = 1}^{N-1} \mu(E_n)$$ for all $N \in \Bbb N$. Now $\mu(E_0) = \mu([a,b]) < \infty$ and $$\int_0^N \mu(E_\alpha)\, d\alpha = \int_0^N \int_a^b 1_{E_\alpha}\, dx\, d\alpha = \int_a^b \int_0^N 1_{E_\alpha}\, d\alpha\, dx = \int_a^b \min\{f(x),N\}\, dx$$ for all $N \in \Bbb N$. Therefore, $\sum_{n = 1}^\infty \mu(E_n) < \infty$ if and only if $\lim_{N \to \infty} \int_a^b \min\{f(x), N\}\, dx$ is finite (the limit already exists since $f_n(x):= \min\{f(x),N\}$ is an increasing sequence of nonnegative measurable functions on $[a,b]$). By the monotone convergence theorem, $\lim_{N \to \infty} \int_a^b \min\{f(x),N\}\, dx = \int_a^b f(x)\, dx$. Thus, $\sum_{n = 1}^\infty \mu(E_n) < \infty$ if and only if $\int_a^b f(x)\, dx < \infty$.